# Why doesn't the "zig-zag" comb deformation retract onto a point, even though it's contractible?

I am starting to read Hatcher's book on Algebraic Topology, and I am a little stuck with exercise 6(c) in Chapter $0$. Unfortunately a picture is involved so it doesn't quite make sense for me to repeat it here - but I'll do it anyways (maybe it helps)

Let $Z$ be the zigzag subspace of $Y$ homeomorphic to $\mathbb{R}$ indicated by the heavier line in the picture: Show there is a deformation retraction in the weak sense of $Y$ onto $Z$, but no true deformation retraction."

Now the definition for deformation retraction and weak deformation retraction as as follows:

We say that $f: X \to X$ is a deformation retraction of a space $X$ onto a subspace $A \subset X$ if there exists a family of maps $f_t : X \to X$ with $t \in [0,1]$ such that $f_0 = \mathbb{1}$ (the identity) and $f_1 (X) = A$ and also $f_t$ restricts to the identity on $A$ for each $t$.

A weak deformation retraction is almost the same, only that we now relax the conditions $f_1(X) = A$ to $f_1 \subset A$ and, for each $t \in [0,1]$ we require that $f_t(A) \subset A$.

Hence, as far as I understand the question and the concepts involved, I need to show that basically no map can be constructed such that the bold zigzag - line stays put while the thin lines retract to the bold line over a finite time interval $0 \leq t \leq 1$. However, I should be able to show that I can pull the thin lines continously onto the thick line, provided I am allowed to move points on the zigzag line .. is that correct ?

Now, the problem is I can't see why I shouldn't be able to do the former, given that I could find a weak deformation retract. Any help would be great!

• For future users, here's a question about proving that it does weakly deformation retract onto a point (i.e. it really is contractible). Commented Jan 11, 2016 at 14:49
• @NajibIdrissi Note that following Hatcher's terminology of "deformation retraction in the weak sense", the space above does not weakly deformation retracts to a point, as that would be equivalent to a strong deformation retract in that case. What is true is that there is a deformation retraction to a point in the sense of Wikipedia, i.e., the deformation retraction need not fix the point throughout the homotopy. This is different from the notions in Hatcher.) Commented Jun 28, 2022 at 23:26

Hint: $Z$, being homeomorphic to $\mathbb{R}$, deformation retracts to a point. Compositions of deformation retractions are deformation retractions (composition in the sense of doing the first deformation retraction for $0\leq t\leq \tfrac{1}{2}$, and doing the second for $\frac{1}{2}\leq t\leq 1$). Thus, if $Y$ deformation retracts to $Z$, it must also deformation retract to a point. Do you see why this is impossible? The argument uses Problem 5 in the same section. The details are in a spoiler box below (put your cursor over it to reveal).

By Problem 5, any neighborhood $U$ of such a point would have to contain an open set $V$ whose inclusion $i:V\hookrightarrow U$ is nullhomotopic; however, this is impossible because any open set $V$ will meet non-path-connected parts of $U$.

• WOW! First, thanks for editing my post and adding the picture! Now I know how to do that so I can post better questions. Secondly - GREAT answer !! It helped a lot working through the hint you gave and then thinking on my own again, with the chance to check against your suggestion in the spoiler box. THANKS !! Commented Mar 30, 2012 at 20:22
• @harlekin: No problem, glad to help! Commented Apr 1, 2012 at 23:03
• That's very very helpful, thank you Zev. Commented Sep 1, 2013 at 19:45

The following argument uses the compactness of $I$ in a way different from that used in the 'lemma' of problem 0.5 of the book.

To show that the entire space $Y$ cannot deformation retract to any point $x\in Y$, consider a sequence of points $x_{n}\rightarrow x,$ with each point $x_{n}$ being in a unique distinct strand (the red dots in the figure are supposed to represent the points of the sequence $x_{n}$).

If $Y$ deformation retracts to $x\in Y$, there is a continuous path from $x_{n}$ to $x$, for each $n\geq 0$. Each such path has to pass through the point $B$ at least once. (if you remove the point $B$ from $Y$, then $x_{n}$ and $x$ are in two distinct path components, hence there cannot exist any path between the two points avoiding $B$).

Given $n$, consider the 'time' $t_{n}$ when $x_{n}$ is first mapped to the point $B$. Then consider the sequence $(x_{n},t_{n})\in Y\times I$. Because $I$ is sequentially compact, we have a converging subsequence $t_{n_{i}}\rightarrow t\in I$. Then $(x_{n_{i}},t_{n_{i}})\rightarrow (x,t)$. Let $F:Y\times I\rightarrow Y$ be the continuous map that gives the deformation retraction. Then the sequence lemma implies $F(x_{n_{i}},t_{n_{i}})\rightarrow F(x,t)$. This is a contradiction as $F(x_{n_{i}},t_{n_{i}})=B \ \forall n_{i}\geq 0$ as constructed, and $F(x,t)=x$, because it is a deformation retraction.

A similar argument can be applied to points $z\in Z$, to show both the facts that there is no def. retraction of $Y$ to $Z$, nor one from $Y$ to $z$.